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How do we get the pH of an acetic acid that is 0.5 M and has a Ka of 2.5 x10^-5?
Rhonda
Rhonda
Ka of an acid
- Thread starter RHONDAROBINSON
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i get 3.5x10^-3 so pH= 2.5ish
CH3COOH + H20--> CH3COO- + H30
Ka=[CH3COO-][H30]/[CH3COO-]
2.5x10^-5=(x)(x)/[0.5]
x=3.5x10^-3 pH=-log[H3O]
a shortcut for monoprotic acids is multiplying the Ka by the acid concentration, then taking the square root. hope this helps
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I am not seeing how you got this pH. Did you multiply 2.5x10 ^5 and 0.5?
Rhonda
Rhonda
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yes multiply 2.5x10^-5 and 5.0x10^-1 = 12.5x10^-6 = (x)(x)
square root each side. x= 3.5x10^-3
this is the concentration of [H3O] so we take the -log (3.5x10^-3)=2.6ish. this is the pH.
check my math, could be wrong.
with pH and pOH just estimate the logs.
square root each side. x= 3.5x10^-3
this is the concentration of [H3O] so we take the -log (3.5x10^-3)=2.6ish. this is the pH.
check my math, could be wrong.
with pH and pOH just estimate the logs.
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just to clarify in case someone is unsure of how to estimate the logs, you start off with the "10^x"
meaning if they ask you to calculate:
pH of 1 * 10^1, then ph = 1
if it's 1 * 10^2, then ph = 2
if it's 1 * 10^3 , then ph = 3
notice how with all these, it's always 1 multiplied by the logs, therefore the pH is whatever the log is being raised to
knowing that trick, if now however you make it 2 * 10^3, then you notice that the pH is about 2.7. Meaning you start off with, in this case 3 and you decrease it based on how big that value that you're multiplying by.
best way to practice this is to just whip out the calculator and try out a couple problems and you'll notice the trend.
I apologize if description was difficult to understand. Did my best to explain math using english haha. Good luck
meaning if they ask you to calculate:
pH of 1 * 10^1, then ph = 1
if it's 1 * 10^2, then ph = 2
if it's 1 * 10^3 , then ph = 3
notice how with all these, it's always 1 multiplied by the logs, therefore the pH is whatever the log is being raised to
knowing that trick, if now however you make it 2 * 10^3, then you notice that the pH is about 2.7. Meaning you start off with, in this case 3 and you decrease it based on how big that value that you're multiplying by.
best way to practice this is to just whip out the calculator and try out a couple problems and you'll notice the trend.
I apologize if description was difficult to understand. Did my best to explain math using english haha. Good luck
it's negative log, btw.
And to clarify the math found in other posts-
sqrt(a x b^c) = sqrt(a) x sqrt (b^c). To use this most effectively you want "a" to be a number you can roughly estimate the square root of, and you want "c" to be an even number (so that you can cleanly divide the exponent by 2). sqrt(1.6x10^-5) = sqrt(16x10^-6) = sqrt(16) x sqrt (10^-6) = 4x10^-3
Next, say you start with 1x10^-3. As you progress further up in value, 3x10^-3, 6x10^-3, you approach 10X10^-3 or 1x10^-2. So anything higher than 1x10^-3 but lower than 1x10^-2 will have a ph between 3 and 2.
-log (4x10^-3) ~~ 2.4ish
also keep in mind that the actual setup should be:
2.5x10^-5=(x)(x)/[0.5 - x]
However with the assumption that you're dealing with a weak acid, x should be sufficiently small that you assume it's negligible compared to 0.5. If you deal with a slightly stronger acid, you have to start considering the x value. For true strong acids the denominator will be so small and Ka so large that you just take [H+] to be the normality of the acid.
And to clarify the math found in other posts-
sqrt(a x b^c) = sqrt(a) x sqrt (b^c). To use this most effectively you want "a" to be a number you can roughly estimate the square root of, and you want "c" to be an even number (so that you can cleanly divide the exponent by 2). sqrt(1.6x10^-5) = sqrt(16x10^-6) = sqrt(16) x sqrt (10^-6) = 4x10^-3
Next, say you start with 1x10^-3. As you progress further up in value, 3x10^-3, 6x10^-3, you approach 10X10^-3 or 1x10^-2. So anything higher than 1x10^-3 but lower than 1x10^-2 will have a ph between 3 and 2.
-log (4x10^-3) ~~ 2.4ish
also keep in mind that the actual setup should be:
2.5x10^-5=(x)(x)/[0.5 - x]
However with the assumption that you're dealing with a weak acid, x should be sufficiently small that you assume it's negligible compared to 0.5. If you deal with a slightly stronger acid, you have to start considering the x value. For true strong acids the denominator will be so small and Ka so large that you just take [H+] to be the normality of the acid.
